and P will be greater than one to R and half AE, and of courſe, AQ

(a fourth proportional to S, R and P) greater than a third proportional

to P and half AE; in which caſe the rectangle AQ, P will be greater

than the ſquare on half AE, and ſo AD (a mean proportional between

AQ and P) greater than half AE; but when this happens, it is plain that

DH can neither cut nor touch the circle on AE, and therefore, the

problem becomes impoſſible.

##
73.
PROBLEM III. (Fig. 4. and 5.)

In any indefinite ſtraight line let there be aſſigned the points A and E; it is required to cut it in another point O, ſo that the ſquare on the ſegment

AO may be to the rectangle contained by the ſegment EO and a given line

P, in the ratio of two given ſtraight lines R and S.

Analysis
. Suppoſe the thing done, and that O is the point ſought: then will the ſquare on AO be to the rectangle EO, P as R to S. Make

AQ to P as R is to S; then will the ſquare on AO be to the rectangle EO,

P as AQ is to P; or (
Eu
. V. 15.) the ſquare on AO is to the rectangle EO,

P as the rectangle AQ, AO is to the rectangle P, AO; wherefore AO is to

EO as AQ to AO; conſequently by compoſition, or diviſion, AO is to AE

as AQ is to OQ, and ſo (
Eu
. VI. 16.) the rectangle AO, OQ is equal to

the rectangle AE, AQ; and hence, as the ſum or difference of AO and OQ

is alſo given, theſe lines themſelves are given by the 85th or 86th of the

Data.

Synthesis
. Take AQ a fourth proportional to S, P and R, and

deſcribe thereon a circle; erect at A, the indefinite perpendicular AK, and

take therein AD, a mean proportional between AE and AQ; from D,

draw DH, parallel to AE, if O be required beyond E; but through F the

center of the circle on AQ, if it be ſought beyond A, or between A and

E, cutting the ſaid circle in H: Laſtly, from H draw HO perpendicular

to DH, which will cut the indefinite line in O, the point required.

For it is plain from the Conſtruction, that AD and HO are equal; and

(
Eu
. VI. 17) the rectangle AE, AQ is equal to the ſquare on AD, and

therefore equal to the ſquare on HO; but the ſquare on HO is equal to the

rectangle AO, OQ, (
Eu
. III. 35. 36) conſequently the rectangle AO, OQ